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Condensed - HW 10 :: Superfluid

PHSX 545

Problem 1 Using the -roton model of 4He spectrum, find the entropy, heat capacity and density of the normal component at low temperature.

Problem 2 (a) Find the E and the angular L carried by a single vortex placed along the rotation axis of p inφ a long cylinder of radius R. Take the superfluid wave function in the form ψ(r, φ) = N/V e θ(r − rc) with the uniform superfluid density ρs (n = 0, 1, 2,... and θ(x) is the step-function). From this deduce the critical angular frequency ωc for nucleation of a single vortex (when Erot = E − Lω becomes negative). (b) A star of radius R = 10km rotates with frequency Ω/2π = 1 Hz. Find the area density nv of vortices in neutron superfluid, assuming the rotational motion of the star is due to vortices, their density is uniform, and the single-quantized vortices are along the rotation axis. Hint: calculate the circulation along a circle of radius r inside the star. 2

Answer of exercise 1 The and distribution functions of the and rotons in the reference frame of (non-moving) condensate P ˆ†ˆ are obtained from the Hamiltonian H = p6=0 εp bpbp with: 1 εph(p) = up nph(p) = eβup−1 2 (p−p0) −βεr (p) εr(p) = ∆ + nr(p) ≈ e 2mr The partition function and free energy of the excitations is 1  −βH Y X −βεp  X X Z = Tr e = ⇒ F = −T ln Z = T ln 1 − e = − εp − T ln np 1 − e−βεp p6=0 p6=0 p6=0 p6=0 The entropy carried by the can be brought to the standard form using the Bose distribution function, ∂F X S = − = [(n + 1) ln(n + 1) − n ln n ] ∂T p p p p p6=0 and from it we get the heat capacity:   ∂S X ∂np ∂E C = T = ε = v ∂T p ∂T ∂T v p6=0 We can calculate entropy directly, or we can start by calculating the energy. The phonon contribution: Z d3k uk u Z ∞ k3dk T 4 Z ∞ x3dx T 4 Energy: Eph = 3 βuk = 2 3 βuk = 2 3 3 x = 2 3 3 Γ(4) ζ(4) (2π~) e − 1 2π ~ 0 e − 1 2π ~ u 0 e − 1 2π ~ u |{z} |{z} 3! π4/90

∂E 3 4Γ(4)ζ(4) 3 Specific heat: Cph = = T - now agrees with experimental T ! ∂T 2π2~3u3

Z C 1 3 4Γ(4)ζ(4) Cph Entropy: Sph = dT = T = T 3 2π2~3u3 3

The roton contribution: ∆  T and np  1: ∆ ∞ Z 3  2  (k−p )2 − Z  2  (k−p )2 d k (k − p0) − ∆ − 0 e T 2 (k − p0) − 0 T 2mr T 2mr T Energy: Er = 3 ∆ + e = 2 3 k dk ∆ + e (2π~) 2mr 2π ~ 0 2mr

∆ √ − Z ∞  2  2 2 e T 2 k − k p0∆ 2πmrT − ∆ 2mr T T = 2 3 (k + p0) dk ∆ + e ≈ 2 3 e 2π ~ −∞ 2mr 2π ~ √ 2   ∂E p0∆ 2πmrT 1 ∆ − ∆ Specific heat: Cr = = + e T ∂T 2π2~3 2T T 2

  −∆/T 2 Z ∞ 2 X X εr(p) ∆ e p0 − (p−p0) Entropy directly from definition: S ≈ [n − n ln n ] = n 1 + ≈ dpe 2mr T r p p p p T T 2π2 3 p6=0 p6=0 ~ 0 √ p2∆ 2m πT = 0 r e−∆/T 2π2~3 T The density of normal component (quasiparticles) for Bose is 4 ∞  T Z 3   Z βεp  4Γ(4)ζ(4) , phonons 1 d p 2 dnp 1 1 4 e 1 1 u5√ ρn(T ) = 3 p − = 2 3 p dp 2 = 2 3 4 3 (2π ) dε 3 2π T βεp 3 2π p 2πmrT ~ p ~ (e − 1) ~  0 e−∆/T , rotons 0 T 3

Answer of exercise 2 (a) The energy of a vortex line with phase of the order parameter θ(r) = nφ in the center of the cylindrical vessel of height H is

Z 2π Z R 2 ˆ ρsvs ~ ~n φ E = H dφ rdr vs = ∇θ = 0 rc 2 m m r that gives for energy of a single vortex

2 2 ρsn ~ R E1 = H2π 2 ln 2m rc

The angular momentum of a vortex is

Z Z R 2 ~nR L1 = H dφ rdr (rρsvs) = 2πHρs 0 2m and the energy in the rotating reference frame

Erot = E − Lω becomes negative when a single vortex nucleation is favorable:

E1 ~n R ω > ωc = = 2 ln L1 mR rc

(b) We can write the velocity field due to many vortices as a sum over individual vortices at positions ri:

X ∇ × v (r) = 2π ~ δ(r − r ) s m i i

The mass here is the mass of particles making up Bose condensate. In neutron star this is pairs of (2 making up a Bose ), so m = 2mn. The circulation along circle of radius r is integration over area of the contour: I ZZ ~ vs · dr = (∇ × vs) · dS = 2π NC C m

2 where NC is the number of vortices inside the circle, NC = nvπr . Assuming azimuthal symmetry the contour integral on the left is 2πrvs(r) and the velocity of the superfluid flow as a function of the radius is

n πr v (r) = ~ v φˆ - just like solid body rotation ωr s m 2 and thus we have πn hn 4mΩ 4 ∗ 2 ∗ 1.7 10−27kg ∗ 1 Hz Ω = ~ v = v ⇒ n = = ∼ 108 1/m2 2m 4m v h 10−34 m2 kg Hz or 10,000 per square centimeter. Vortex separation distance is about 0.1 mm.